The first thing I notice is that there is cyclic symmetry between $x$ and $y$, and so setting $y=x$, we find:
$$2x^3+3x^2-1=(x+1)^2(2x-1)=0$$
Thus, we know the points:
$$(x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)$$
are on the given curve. Next, if we begin with the line:
$$y=1-x$$
and cube both sides, we obtain:
$$y^3=1-3x+3x^2-x^3$$
We may arrange this as:
$$x^3+3x(1-x)+y^3=1$$
Since $y=1-x$, we may now write
$$x^3+3xy+y^3=1$$
And since the point $$\left(\frac{1}{2},\frac{1}{2} \right)$$ is on the line $y=1-x$, we know the locus of the given curve is the line $y=1-x$ and the point $(-1,-1)$. Hence, there can only be one set of points on the given curve that are the vertices of any triangle, equilateral or otherwise.
Using the formula for the distance between a point and a line, we find the altitude of the equilateral triangle will be:
$$h=\frac{|(-1)(-1)+1-(-1)|}{\sqrt{(-1)^2+1}}=\frac{3}{\sqrt{2}}$$
Using the Pythagorean theorem, we find that the side lengths of the triangle must be:
$$s=\frac{2}{\sqrt{3}}h=\sqrt{6}$$
And so the area of the triangle is:
$$A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}$$
